An aqueous mixture containing starch (a colloid), NaCl, glucose, and albumin (a colloid) is placed in a dialyzing bag and immersed in distilled water. Which of the following correctly describes the location of the indicated substance after dialysis?
A-albumin, inside
B-starch outside
C-albumin inside and outside
D-water inside only
E-starch inside and outside
Answer: A Correct
Explanation: A dialyzing bag is a visking device i.e. artificial semi-permeable membrane tubing used for a differential molecular movement across to the medium/cell membrane (separation techniques), that allows the flow of smaller molecules (low-molecular-weight molecules) in solution on the basis of differential diffusion.
Tiny molecules like water and glucose can pass through its microscopic pores while larger molecules like salt ions of NaCl, albumin, sucrose and starch cannot pass through the pore.
So, the salt ions of NaCl, albumin, sucrose and starch stays inside the bag, only glucose can go out of the bag and water may also diffuse into the bag from outside .
After dialysis, starch will be found outside the bag, while glucose, albumin, and NaCl will be found inside the bag.
An aqueous mixture containing starch (a colloid), NaCl, glucose, and albumin (a colloid) is placed in a dialyzing bag and in distilled water. After dialysis, the starch will be found outside the bag, while the glucose, albumin, and NaCl will be found inside the bag. The starch particles are too large to pass through the semi-permeable dialyzing membrane, so they remain outside the bag, while the smaller glucose, albumin, and NaCl molecules are able to pass through the membrane and are found inside the bag.
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In the quantum mechanical model of the atom, electrons are said to occupy three dimensional regions of probability around the nucleus called Each of these is assigned a to indicate its relative size and and energy:________
Answer:
Although the question is not clear enough, I'll explain in the most helpful way that I can.
Explanation: The Quantum Mechanical Model of the atom is one of two models that explains the structure of an atom. In this model, it is imposssible to know the exact position and momentum of an electron at the same time and thus its position is uncertain. However, electrons are said to occupy three dimensional regions of probability referred to as 'ORBITALS'. You can simply refer to the orbitals as where there is very high likelihood for one or two electrons to be found. Each of these orbials are depicted with a number, n and a letter(s,p,d,f or g) eg 1s, 2p, etc. The 'number' signifies the distance of the orbital from the nucleus and the energy level of the electron in an atom while the 'letter' indicates the shape of the orbital, for example, the s-orbital is spherical in shape.
I hope this helps!
The following statements about blood are true except that __________.
Except for __________, the following statements about blood are true.
a.The viscosity is three to five times greater than water.
b The pH is slightly acidic.
c. It contains about 55% plasma.
D. It contains dissolved gases.
Answer:Except for the pH is slightly acidic, the following statements about blood are true.
Explanation:Viscosity is an inherent feature of liquid compared to the inner resistance of nearby fluid films pushing through one another. All blood has a much greater viscosity than water. Authorities regard these matters as “non-Newtonian fluids,” of which ketchup and blood are excellent illustrations.
Blood is usually slightly basic, with a typical pH scale of approximately 7.35 to 7.45. Plasma composes 55% of the whole blood volume. The albumin included in plasma restricts the blood from dropping too much water. The plasma of vertebrates also includes dissolved gases.
Blood is a fluid connective tissue that is 92 percent water. It is slightly more acidic, viscous, and salty than water.
Explanation:Blood is a fluid connective tissue composed of plasma, dissolved substances, and blood cells. It is about 92 percent water, making the statement provided true. However, blood is slightly more acidic than water, slightly more viscous than water, and slightly more salty than seawater, making these statements false. Red blood cells carry oxygen, white blood cells defend the body, and platelets help blood clot.
Question 1: A substance that is soluble in two liquids and makes an emulsion last longer is called what?
Question 2: What is the process called that reduces the size of particles so emulsions will last longer?
Answer:
1. A substance that is soluble in two liquids and makes an emulsion last longer is called "Emulsifier".
2. The process that reduces the size of particles so emulsions will last longer is called "Homogenization".
Explanation:
Emulsifiers are additives which enable two liquids to mix around each other. Water and oil separate in a container, for an instance, but using an emulsifier can make the liquids blend along. It is widely used on various foods and beverages. Egg yolks and mustard are a few examples of emulsifiers.
Homogenization is the physical mechanism by which the fat molecules in milk are broken down because then they stay incorporated instead of segregated as cream. Majority of the milk sold in the United States is homogenized.
A BaSO4 slurry is ingested before the gastrointestinal tract is x-rayed because it is opaque to x-rays and defines the contours of the tract. Ba2+ ion is toxic, but the compound is nearly insoluble. If ΔG o at 37°C (body temperature) is 59.1 kJ/mol for the process BaSO4(s) ⇌ Ba2+(aq) + SO42−(aq) what is the [Ba2+] in the intestinal tract? (Assume that the only source of SO42− is the ingested slurry.)
Explanation:
The given reaction equation is as follows.
[tex]BaSO_{4}(s) \rightleftharpoons Ba^{2+}(aq) + SO_{4}^{2-}(aq)[/tex]
The value of [tex]\Delta G^{o}[/tex] = 59.1 kJ/mol
We know that ,
[tex]\Delta G^{o} = -RT ln K_{sp}[/tex]
or, [tex]ln K_{sp} = -(\frac{\Delta G^{o}}{RT}) [/tex]
= -(\frac{59.1 kJ/mol}{(8.314 \times 10^{-3} kJ/mol.K \times 310 K))}[/tex]
= -22.93
or, [tex]K_{sp} = e^{-22.93}[/tex]
= [tex]1.1 \times 10^{-10}[/tex]
[tex]K_{sp} = [Ba^{2+}][ SO_{4}^{2-}][/tex]
Therefore, [tex][Ba^{2+}] =\sqrt{K_{sp}}[/tex]
= [tex]\sqrt{ 1.1 \times 10^{-10}}[/tex]
= [tex]1.05 \times 10^{-5} M[/tex]
Therefore, we can conclude that the value of [tex][Ba^{2+}][/tex] in the intestinal tract is [tex]1.05 \times 10^{-5} M[/tex].
The concentration of Ba2+ (barium ions) in the intestinal tract following the ingestion of BaSO4 slurry can be calculated using the equilibrium constant (K) derived from the Gibbs free energy equation. After performing the calculations, [Ba2+] is found to be 3.45 x 10⁻⁶ M.
Explanation:The question is essentially asking for the concentration of Ba2+ ions in the intestinal tract following the ingestion of BaSO4 slurry, given a ΔG° value of 59.1 kJ/mol at 37°C. In order to solve this, we need to consider the process of the reaction: BaSO4(s) ⇌ Ba2+(aq) + SO42−(aq). From the Gibbs free energy equation, we can calculate the equilibrium constant (K) as: K = e^(-ΔG°/RT), where R is the gas constant (R = 8.314 J/molK) and T is the temperature in Kelvin (37°C = 310K).
Therefore, ΔG° = -RTlnK -> K=e^(-ΔG°/RT). We find K = e^(-(−59.1×10³J mol⁻¹)/((8.314 J K⁻¹ mol⁻¹)(310 K))), which gives K = 1.19 x 10⁻¹⁰. Since the equilibrium involves the dissolution of one BaSO4 molecule to give one Ba2+ ion and one SO42- ion, the equilibrium concentrations of Ba2+ and SO42- are equal. Hence, [Ba2+] = sqrt(K) = sqrt(1.19 x 10⁻¹⁰) = 3.45 x 10⁻⁶ M.
In conclusion, the concentration of Barium ions, Ba2+, in the intestinal tract following ingestion of BaSO4 slurry would be around 3.45 x 10⁻⁶M.
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Crime scene investigators keep a wide variety of compounds on hand to help with identifying unknown substances they find in the course of their duties. One such investigator, while reorganizing their shelves, has mixed up several small vials and is unsure about the identity of a certain powder. Elemental analysis of the compound reveals that it is 63.57% carbon, 6.000% hydrogen, 9.267% nitrogen, 21.17% oxygen by mass. Which of the following compounds could the powder be?
a.) C11H15NO2 = 3,4-methylenedioxymethamphetamine (MDMA), illicit drug
b.) C3H6NO3 = hexamethylene triperoxide diamine (HMTD), commonly used explosive
c.) C21H23NO5 = heroin, illicit drug
d.) C8H9NO2 = acetaminophen, analgesic
e.) C7H5N3O6 = 2,4,6-trinitrotoluene (TNT), common used explosive
f.) C17H19NO3 = morphine, analgesic
g.)C10H15N = methamphetamine, stimulant
h.) C4H5N2O = caffeine, stimulant
Answer:
Option d: C₈H₉NO₂ = acetaminophen, analgesic
Explanation:
% composition of compound is:
63.57 g of C
6 g of H
9.267 g of N
21.17 g of O
First of all we divide each by the molar mass of the element
63.57 g / 12 gmol = 5.29 mol of C
6 g of H / 1 g/mol = 6 mol H
9.267 g of N / 14 g/mol = 0.662 mol of N
21.17 g of O / 16 g/mol = 1.32 mol of O
We divide each by the lowest value, in this case 0.662
5.29 / 0.662 = 8
6 / 0.662 = 9
0.662 / 0.662 = 1
1.32 / 0.662 = 2
Molecular formula of the compound is C₈H₉NO₂
The density of water is 1.00 gram/milliliter. What is the volume in milliliters of 1.00 mole of water? Express your answer to the correct number of significant figures.
Answer:
18.00 mL is the volume in mL of 1 mol of water
Explanation:
Water density = water mass / water volume
1 g/mL = water mass / water volume
1 mol of water weighs 18 g. Therefore, 1 g/mL = 18 g / water volume
water volume = 18 mL
Answer:
18.0
Explanation:
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If x is a string, then x = new String("OH"); and x = "OH"; will accomplish the same thing. Group of answer choices True False
Answer:
True is the correct answer to the above question.
Explanation:
If x is a string then it can be assigned by the help of two ways in java:By the help of constructor:- When we write " x = new String("OH");", then it will create a pass a string "OH" into the constructor. It is because the String is a class in java and x is an object created by the constructor of the String class.With the help of assigning: The "x= OH", which assigns the value of x which is an object of String class it can also use the constructor to initialize the "OH" string on the class.The above question states the two scenarios which are defined above. Hence the question statement is true.Answer:
"True" is the correct answer to this question.
Explanation:
The program to the given question as follows:
Program:
public class data //defining class
{
public static void main (String [] aw)//defining the main method
{
String x="OH"; //defining string variable x and assign value
System.out.println("assign value: "+x); //print value
x = new String("OH"); //defining instance variable and assign value
System.out.println("assign value by creating instance: "+x); //print value
}
}
Output:
assign value: OH
assign value by creating instance: OH
Explanation of the program:
In the above java program, a class data is defined, inside the class the main method is declared, In this main method a string variable "x" is defined that holds a value "OH", then we the print function to print this variable value.
In the next line, An instance of variable x is created, which holds a value "OH" in its parameter. In this question, both are correct because both hold the same value.
A stock solution is prepared by dissolving 12.5 g of NaCI in enough water to prepare 150.0 mL of solution. What volume of this stock solution will be used to prepare a diluted solution that is 250.0 mL of a 0.500 M solution of NaCI?
Answer:
8.75 mL
Explanation:
First, we calculate the molar mass of NaCl = molar mass of Na + molar mass of Cl. Molar mass of Na = 23 g/mol, molar mass of Cl = 35.5 g/mol.
So molar mass NaCl = (23 + 35.5) g/mol = 58.5 g/mol. The number of moles ,n of NaCl in 12.5g is n = mass of NaCl/ molar mass NaCl = 12.5 g / 58.5 g/mol = 0.214 mol.
The molarity, M of 150 mL M = number of moles/ volume = 0.214 mol / 150 mL = 1.427 M.
We now calculate the number of moles of NaCl in 250 mL of 0.500 M.
Number of moles, n = molarity × volume. molarity = 0.500 M, volume = 250 mL. So n = 0.500 × 250 = 0.125 moles. Since we have 0.125 moles in the dilute 250 mL solution, the volume of the 150 mL 1.43 M solution required is number of moles in 250 mL solution/molarity of 150 mL solution = 0.125 mol / 1.427 M = 0.0875 L = 8.75 mL
When a current is passed through a solution of salt water, sodium chloride decomposes according to the following reaction: NaCl + H2O → NaOH + Cl2 + H2 Balance the equation. Choose "blank" if no coefficient is needed. NaCl + H2O → NaOH + Cl2 ++ H2
Answer:
The answer to your question is 2NaCl + 2H₂O ⇒ 2NaOH + Cl₂ + H₂
Explanation:
Original chemical equation
NaCl + H₂O ⇒ NaOH + Cl₂ + H₂
Reactant Element Products
1 Na 1
1 Cl 2
2 H 3
1 O 1
This reactions is unbalanced
2NaCl + 2H₂O ⇒ 2NaOH + Cl₂ + H₂
Reactant Element Products
2 Na 1
2 Cl 2
4 H 4
2 O 2
Now, the reaction is balanced
Answer:
NaCl + 2 H2 O → 2 NaOH + Blank Cl 2 + Blank H 2
describe how these nutrients and elements such as carbon, oxygen, and nitrogen are cycled through ecosystems.
Answer:
Carbon Cycle
Steps of the Carbon Cycle
CO2 is removed from the atmosphere by photosynthetic organisms (plants, cyanobacteria, etc.) and used to generate organic molecules and build biological mass. Animals consume the photosynthetic organisms and acquire the carbon stored within the producers. CO2 is returned to the atmosphere via respiration in all living organisms. Decomposers break down dead and decaying organic matter and release CO2. Some CO2 is returned to the atmosphere via the burning of organic matter (forest fires). CO2 trapped in rock or fossil fuels can be returned to the atmosphere via erosion, volcanic eruptions, or fossil fuel combustion.Nitrogen Cycle
Steps of the Nitrogen Cycle
Atmospheric nitrogen (N2) is converted to ammonia (NH3) by nitrogen-fixing bacteria in aquatic and soil environments. These organisms use nitrogen to synthesize the biological molecules they need to survive. NH3 is subsequently converted to nitrite and nitrate by bacteria known as nitrifying bacteria. Plants obtain nitrogen from the soil by absorbing ammonium (NH4-) and nitrate through their roots. Nitrate and ammonium are used to produce organic compounds. Nitrogen in its organic form is obtained by animals when they consume plants or animals. Decomposers return NH3 to the soil by decomposing solid waste and dead or decaying matter. Nitrifying bacteria convert NH3 to nitrite and nitrate. Denitrifying bacteria convert nitrite and nitrate to N2, releasing N2 back into the atmosphere.Oxygen Cycle
Oxygen is an element that is essential to biological organisms. The vast majority of atmospheric oxygen (O2) is derived from photosynthesis. Plants and other photosynthetic organisms use CO2, water, and light energy to produce glucose and O2. Glucose is used to synthesize organic molecules, while O2 is released into the atmosphere. Oxygen is removed from the atmosphere through decomposition processes and respiration in living organisms.
Explanation:
A(n) _______ solution has a higher concentration of water and lower concentration of solute than the cell placed in the solution.
Answer : A hypotonic solution has a higher concentration of water and lower concentration of solute than the cell placed in the solution.
Explanation :
Solution : It is made up of the combination of amount solute and solvent.
Isotonic solutions : It is defined as the solutions in which the concentration of solute inside the cell and outside the cell is same.
Hypotonic solutions : It is defined as the solutions in which the concentration of solute inside the cell is lower than outside the cell.
For example : Diluted sugar syrup
Hypertonic solutions : It is defined as the solutions in which the concentration of solute inside the cell is higher than outside the cell.
For example : Concentrated sugar syrup
Hence, a hypotonic solution has a higher concentration of water and lower concentration of solute than the cell placed in the solution.
The term for a solution that has a higher concentration of water and a lower concentration of solute than a cell is 'hypotonic'. In this scenario, water moves into the cell via osmosis.
Explanation:A(n) hypotonic solution has a higher concentration of water and lower concentration of solute than the cell placed in the solution. In biology, we often talk about the relationship between cells and their surrounding environment in terms of tonicity. In a hypotonic environment, there is less solute (like salt or sugar) outside the cell compared to inside the cell. This causes water to move into the cell by osmosis, because water moves from areas of high concentration to areas of low concentration until equilibrium is reached.
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In a sealed gas-liquid system at constant temperature, eventually...?
there will be no more evaporation.
the rate of evaporation equals the rate of condensation.
the rate of condensation decreases to zero.
the rate of condensation exceeds the rate of evaporation.
Explanation:
Evaporation
It is the process of converting liquid into vapors .
Condensation
It is the process of converting vapors back into liquid state .
Suppose if we have a sealed container and we are supplying it with no or little heat , we will see that as we increase heat , the particles starts moving faster .When they move they also colloide and transfer energies .The kinetic energies of certain molecule increase to an extent that they leave the other particles and escape in atmosphere .That is evaporation occurs .At the same time when these vapors collide with each other or with the walls of container they get cooled and again get converted to liquid state .It is seen that a equilibrium is reached when "rate of evaporation becomes equal to rate of condensation ".
For the procedural error, indicate if the error will affect the actual yield of copper(II) saccharinate product and if it does, will it raise or lower the actual yield:__________
Washing the crystals with hot water
Answer:
Lowers the actual yield
Explanation:
Calculate the work (kJ) done during a reaction in which the internal volume contracts from 85 L to 12 L against an outside pressure of 2.4 atm.
Answer:
The work done on the system is 17.75_KJ
Explanation:
To solve this question we need to know the required equations from the given variables, thus
Initial volume = 85L
Final volume = 12L
External pressure = 2.4 atm.
work done = - PΔV
2.4×(85-12) = 175.2L×atm
Converting from L•atm to KJ is given by
1 L•atm = 0.1013 kJ
(175.2 L•atm) * (0.1013 kJ / 1 L•atm)
= 17.75 kJ
The work done on the system is 17.75_KJ
The work done on the system is -17.75 kJ.
Work done is obtained using the relation;
w = PΔV
where;
w = work done
V = change in volume
V1 = 85 L
V2 = 12 L
P = 2.4 atm
Hence;
w = 2.4 atm(12 - 85) L = -175.2 atm L
Now;
1 L atm = 101.325 J
-175.2 atm L = -175.2 atm L × 101.325 J/1 L atm = -17.75 kJ
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A syringe containing 1.64 mL of oxygen gas is cooled from 94.6 ∘C to 0.3 ∘C. What is the final volume Vf of oxygen gas?
Answer:
[tex]V_f=V_2=1.22\times 10^{-3} L=1.22mL[/tex]
Explanation:
Use Charles's law which states: "At constant pressure, the volume occupied by a gas sample is directly proportional to the absolute temperatures they support."
This law can be expressed mathematically as follows:
[tex]\frac{V_1}{T_1}=\frac{V_2}{T_2}[/tex]
Where:
[tex]V_1=Initial\hspace{3}volume=1.64mL=1.64\times 10^{-3} L\\V_2=Final\hspace{3}volume\\T_1=Initial\hspace{3}temperature=94.6$^{\circ}$C=367.75K\\T_2=Final\hspace{3}temperature=0.3$^{\circ}$C=273.45K[/tex]
Solving for [tex]V_2[/tex]
[tex]V_2=\frac{V_1*T_2}{T_1} =\frac{(1.64\times 10^{-3} )*273.45}{367.75} =1.21946431\times 10^{-3} \approx1.22\times 10^{-3} L =1.22mL[/tex]
Silica, sio2, is formed on silicon as an electrically insulating layer for microelectronic devices. silica is formed when silicon is exposed to o2 gas at an elevated temperature. at 900˚c, it takes 90 minutes for the oxygen to diffuse from the surface to form a 0.06 micron (0.06 x 10-6 m) thick layer of sio2 on
Final answer:
Silicon dioxide, or silica, is integral to microelectronics as an electrical insulator and is used in applications such as isolation, gate insulation, and dopant diffusion. Chemical Vapor Deposition is commonly used to deposit silica thin films. The material's tetrahedral structure ensures stability under rapid temperature changes, vital for semiconductor industries.
Explanation:
Silicon dioxide (SiO2), often referred to as silica, plays a crucial role in the world of microelectronics. Its excellent electrical insulating properties make it essential for various applications within semiconductor devices. Silicon dioxide is employed for the isolation of conductive layers as well as for its dielectric properties which are used in gate insulation. Moreover, it serves to facilitate the diffusion of dopants from oxides and as a means to prevent the loss of dopants when capping films.
The Chemical Vapor Deposition (CVD) method is widely used for depositing thin layers of SiO2 during semiconductor processing. This is due to the unique challenges associated with its application in creating insulating thin films. Additionally, silica's unique diamond-like network structure allows for rapid temperature changes, making it invaluable in the steel, electronic, and semiconductor industries.
Silica's three-dimensional tetrahedral structure, where silicon atoms are bonded to oxygen, confers stability and resilience that is crucial for the high temperature processes involved in microelectronic device fabrication. Different forms of silicon dioxide, such as quartz and fused silica, contribute to the diversity of silica's properties and applications. Its naturally abundant presence and the numerous crystalline forms it can take, underscore the material's importance to technology and industry.
Aluminum reacts with excess hydrochloric acid to form aqueous aluminum chloride and 190.6 ml of hydrogen gas (collected over water at 26°C and 0.89 atm). (Vapor pressure of water at 26ºC = 25.2 mmHg.) How many grams of aluminum reacted? Enter to 4 decimal places.
Answer:0.119g
Explanation:equation of rxn is
2Al+6HCl=2AlCl3+3H2
From ideal gas eqn
PV=nRT
n=PV/RT
P here is the partial pressure of H2 from the qtn.According to Dalton law of partial pressure, PT=PH2+PH20
PT=0.89atm given
PH20=25.2mmhg given=25.2/760atm,=0.033atm
PH2=PT-PH20
PH2=0.89-0.033=0.857atm
T=26+273=299K
R=0.082atmdm^-3mol^-1K^-1
V=190.6ml=190.6cm3=190.6/1000=0.1906dm3
n=PV/RT
n=0.857*0.1906/0.082*299
=0.00667moles of H2.
From the eqn of reaction,
2moles of Al reacts to gv 3moles of H2
xmoles of Al will give 0.00667moles of H2
xmoles=0.00667*2/3 (cross multiplying)=0.00444moles of Al
From the relationship, n=mass/MW
mass=MW*n
MW of Al=27g/mol
mass=0.0044moles*27g/moles
mass=0.119grams of Al.
The mass of aluminum that has reacted is 0.119 g. The mass of the reactant can be calculated by finding its moles.
How to calculate the mass of the reactant?the mass of the reactant can be calculated by finding its moles in the reaction and putting the value in the mole formula.
The given reaction is:
[tex]\bold {2Al+6HCl\rightarrow 2AlCl_3+3H_2}[/tex]
First, calculate the moles of the Hydrogen from the ideal gas equation,
[tex]n=\dfrac {0.857\times 0.1906}{0.082\times 299}\\\\ n = \rm 0.00667 \ moles \ of \ H2.[/tex]
The molar ratio between Al and [tex]\bold {H_2 }[/tex] is 3:2.
Thus, moles of Aluminium is:
[tex]\text{Moles of Al} = 0.00667\times \dfrac 23 \\\\\text{Moles of Al} = 0.00444 \rm \ moles[/tex]
Thus the mass of Aluminium,
[tex]m ={\rm 0.0044 \ moles\times 27 \ g/moles}\\\\m = 0.119\rm \ g[/tex]
Therefore, the mass of aluminum that has reacted is 0.119 g.
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When a chemical reaction is in equilibrium,
a. the reaction is proceeding at its maximum rate.
b. there is no net change in the amount of substrates or products.
c. the reaction has stopped.
d. there are equivalent amounts of substrates and products.
Answer: b. there is no net change in the amount of substrates or products.
Explanation:
The reactions which do not go on completion and in which the reactant forms product and the products goes back to the reactants simultaneously are known as equilibrium reactions.
Equilibrium state is the state when reactants and products are present but the concentrations does not change with time.
[tex]A\rightleftharpoons B[/tex]
For a chemical equilibrium reaction, equilibrium state is achieved when the rate of forward reaction becomes equals to rate of the backward reaction.
A gas that has a volume of 28 L, a temperature of 45 °C, and an unknown pressure has its volume increased to 34 L and its temperature decreased to 35 °C. If the pressure is measured after the change to be 2.0 atm, what was the original pressure of the gas?A.1.6 atmB.2.5 atmC.3.2 atmD.4.1 atm
Answer:
The answer to your question is letter B. 2.5 atm
Explanation:
Data
P1 = ? P2 = 2 atm
T1 = 45°C = 318°K T2 = 35°C = 308°K
V1 = 28 L V2 = 34 L
Formula
Combined gas law
P1V1/T1 = P2V2/T2
solve for P1
[tex]P1 = \frac{T1P2V2}{V1T2}[/tex]
Substitution
P1 = (318 x 2 x 34) / (28 x 308)
Simplification
P1 = 21624 / 8624
Result
P1 = 2.5 atm
While studying force fields, Ms. Garcia told her students they were about to experiment with magnets. Before beginning the magnet experiments, Ms. Garcia set up this demonstration: three magnets that seemed to hover, one above the other, without touching. Label the poles of the magnets, A, B and C, in this arrangement. A) A north, B south, C south. B) A south, B north, C north. C) A north, B north, C south. D) A, B, and C are all south poles.
Answer:
The answer is A. a north, B south and C south
Explanation:
i just took the test
Answer:
A
Explanation:
USATestprep
Ice Station Bravo near the North Pole launched a helium-filled balloon to check atmospheric conditions. At sea level (1.0 atm) where the balloon was launched, it had a volume of 0.93m^3 . It rose to an altitude of 18000m where the atmospheric pressure dropped to 0.072atm.
What is the volume of the balloon at that altitude assuming that the temperature was the same at sea level?
Answer:
12.9 m³ is the new volume
Explanation:
As the temperature keeps on constant, and the moles of the gas remains constant too, if we decrease the pressure, the volume will increase. If the volume is decreased, pressure will be higher.
The relation is this: P₁ . V₁ = P₂ . V₂
1 atm . 0.93m³ = 0.072 atm . V₂
0.93m³ .atm / 0.072 atm = V₂
V₂ = 12.9 m³
In conclusion and as we said, pressure has highly decreased so volume has highly increased.
When calculating the wattage for a power supply by estimating its needs, how much wattage would you allow for a fan?
Answer:
A fan is a common house appliance which is attached to the ceiling and uses an electric motor to rotate blades or paddles in a circular motion. Ceiling fans help cool a room by moving air which causes evaporative cooling. Fans range in size from 36 inches to 56 inches using 55 to 100 watts, a typical 48 inch ceiling fan will use 75 watts.
where you have
Hours Used Per Day:5
Power Use (Watts): 75
Price (kWh): 0.01
then you have
Cost Per Hour=0.0075
Cost Per Day= 0.0375
Cost Per Month=1.14
Cost Per Year=13.69
kWh Per Day: 0.38
Answer:
5W
Explanation:
Find the molecular formula of a compound that contains 42.56 g of palladium and 0.80 g of hydrogen. The molar mass of the compound is 216.8 g/mol.
Answer:
The answer to your question is Pd₂H₄
Explanation:
Data
mass of palladium = 42.56 g
mass of hydrogen = 0.8 g
Process
1.- Convert the grams of each substance to moles
106 g of Pd ----------------- 1 mol
42.56 g of Pd ------------- x
x = (42.56 x 1)/106
x = 0.402 moles
1 g of H --------------------- 1 mol
0.8 g of H ------------------ x
x = (0.8 x 1)/1
x = 0.8 moles
2.- Divide by the lowest number of moles
Palladium = 0.402/0.402 = 1
Hydrogen = 0.8/0.402 = 1.99 ≈ 2
3.- Write the empirical formula
PdH₂
4.- Calculate the mass of the empirical formula
PdH₂ = 106 + 2 = 108
5.- Divide the molar mass by the molar mass of the empirical formula
216.8/108 = 2
6.- Get the molecular formula
2(PdH₂) = Pd₂H₄
The molecular formula of the compound is determined to be Pd2H4, based on the calculated mole ratio of palladium to hydrogen and considering the given molar mass of the compound.
Explanation:To find the molecular formula of a compound, we first determine the mole ratio between palladium and hydrogen. The molar mass of palladium (Pd) is 106.42 g/mol and the molar mass of hydrogen (H) is 1.008 g/mol. Therefore, 42.56 g of Pd is equivalent to 42.56/106.42 = 0.4 mol, and 0.80 g of H is equivalent to 0.80/1.008 = 0.793 mol. This simplifies to a ratio of 1:2, indicating the empirical formula of the compound is PdH2.
Next, we compare the molar mass of the empirical formula with the molar mass given for the compound. The molar mass of PdH2 is 1.008*2 + 106.42 = 108.436 g/mol. The molar mass given for the compound is 216.8 g/mol. Therefore, the molecular formula is twice the empirical formula, result to be Pd2H4.
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If the density of a 21.71 m (molal) solution of ethanol, C2H5OH, in water is 0.914 g/mL, what is the molarity of ethanol in this solution?
The molarity of the given ethanol solution can be computed using its given molality and the density of the solution. The molality tells us the moles of ethanol per kilogram of water, and the density allows us to convert this to moles per liter, generally yielding a larger molarity for the same solution. Thus, the molarity of ethanol in this solution would be 914 M.
Explanation:To calculate the molarity of the ethanol solution, we first need to know that the definition of molality (m) is the moles of solute (in this case, ethanol) per kilogram of solvent (here, water). Given that the density of the solution, is 0.914 g/mL, we can convert this to kg/L for ease of calculation. So, 0.914 g/mL is equivalent to 914 kg/m^3.
Next, let's take 21.71 molal solution, implying there are 21.71 moles ethanol in 1 kilogram of water. Since the density of the solution is 0.914 g/mL (or 914 kg/m^3), there would be 914 moles of ethanol in 1 m^3 of solution (since 1 L = 1 m^3).
Thus, the molarity (M), which is defined as the moles of solute per liter of solution, of the ethanol in the solution would be 914 M.
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How much energy must be removed from a 125 g sample of benzene (molar mass= 78.11 g/mol) at 425.0 K to liquify the sample and lower the temperature to 335.0 K? The following physical data may be useful.Hvap = 33.9 kJ/molHfus = 9.8 kJ/molCliq = 1.73 J/g CCgas = 1.06 J/g CCsol = 1.51 J/g CTmelting = 279.0 KTboiling = 353.0 Ka. 95.4 kJ
b. 74.4 kJ
c. 38.9 kJ
d. 67.7 kJ
e. 54.3 kJ
Answer: d) 67.7KJ
Explanation:
Number of moles of Benzene= 125/78.11=1.60 moles
Q = mc◇T = 125× 1.60×(425-353)= 9540J
Q= ◇Hvap× 1.60=
33.9 × 1.60 = 54.24KJ
Q = mc◇T= 125 × 1.73× (353-335)=3803KJ
Total energy required to remove Carbon= (9540 + 54240+ 3893) J =67,673J =67
7KJ
Liquid nitrogen has a density of 0.808 g/mL and boils at 77 K. Researchers often purchase liquid nitrogen in insulated 195-L tanks. The liquid vaporizes quickly to gaseous nitrogen (which has a density of 1.15 g/L at room temperature and atmospheric pressure) when the liquid is removed from the tank. Suppose that all 195 L of liquid nitrogen in a tank accidentally vaporized in a lab that measured 10.00 m× 10.00 m× 2.50 m.What maximum fraction of the air in the room could be displaced bythe gaseous nitrogen?
Answer:
The nswer to the question is
The maximum fraction of the air in the room that could be displaced by the gaseous nitrogen is 0.548 or 54.8 %
Explanation:
To solve the question we note that
The density of the liquid nitrogen = 0.808g/mL and the volume is 195 L tank (vaporised)
Therefore since density = mass/volume we have
mass = Density × volume = 0.808 g/mL × 195 L × 1000 ml/L =157560 g
In gaseous form the liquid nitrogen density =1.15 g/L
That is density = mass/volume and volume = mass/density = 157560 g/(1.15g/L) or
volume = 137008.69565 L
The dimension of the room = 10 m × 10 m × 2.5 m = 250 m³ and
1 m³ is equivalent to 1000 L, therefore 250 m³ = 250 m³ × 1000 L/m³ = 250000L
Therefore fraction of the volume occupied by the gaseous nitrogen =
137008.69565 L/250000 L = 0.548
Therefore the gaseous nitrogen occpies 54.8% of the room
To find the maximum fraction of air displaced by vaporized nitrogen in a room, calculate the mass of liquid nitrogen, convert it to gaseous volume, and compare it to the room's volume.
Explanation:When considering the accidental vaporization of 195 liters of liquid nitrogen in a closed room, we need to calculate the volume of the room and determine what fraction of the room's air could potentially be displaced by the nitrogen gas. To do this, we will use the given room dimensions and the densities of liquid and gaseous nitrogen. The room's volume is 10.00 m x 10.00 m x 2.50 m, which equals 250 cubic meters or 250,000 liters. The density of liquid nitrogen is 0.808 g/mL, and it converts to 0.808 kg/L since there are 1000 milliliters in a liter. Multiplying the density by the total volume of liquid nitrogen gives us the mass in kilograms, which is then converted to the volume of gaseous nitrogen at standard temperature and pressure using the given density of 1.15 g/L.
Now, converting the mass to volume for gaseous nitrogen, we can find out the fraction of room's air displaced by using the ratio of nitrogen gas volume over the room's total volume. This is an application of the ideal gas law where a given mass of gas occupies different volumes based on its phase (liquid or gas) and other conditions such as temperature and pressure.
Which of the following elements form cations (positively charged ions) readily? C, O, Na, Fe, As, Br, K
a. C, O, Na, Fe, As, Br, K
b. C, O, Na
c. Fe, As, Br, K
d. O, Na, Fe
e. Na, Fe, K
Answer:
e. Na, Fe, K
Explanation:
The group of elements that will be positive charge ions is metal. You can find metal in the first 2 columns of the periodic table and in the transition area. Natrium/sodium (Na), iron (Fe), and kalium/potassium(K) categorized as metal and they will form positive charge ions.
On the other hand carbon(C), oxygen (O), arsenic(As) and bromine(Br) is gas and will form negative charge ions. Gas located on the right side of the periodic table.
The elements that readily form positively charged ions, or cations, are more often metals like Sodium (Na), Iron (Fe), and Potassium (K). Therefore, the correct answer from the options given is 'Na, Fe, K'. option e.
Explanation:The elements that form cations, or positively charged ions, readily are elements that tend to lose electrons. This characteristic is typically associated with metals. In the options given, Na (Sodium), Fe (Iron), and K (Potassium) are the ones which are more likely to form cations. So, the correct answer to your question: 'Which of the following elements form cations (positively charged ions) readily: C, O, Na, Fe, As, Br, K?' would be option e. Na, Fe, K.
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Which of the following is the correct ranking of the three bonds and interactions in order from highest to lowest in terms of their bond strength between two side chains of a protein in their tertiary structure?
I. Disulfide bond between two cystines
II. Hydrophobic interactions between two leucines
III. H-bonding in water
Answer:
I > III > II
Explanation:
I) A disulfide bond between two cystines is created when a sulfur atom from one cystine forms a strong, single covalent bond with a sulfur atom from a second cystine. When a disulfide bond is created, each cystine loses one hydrogen atom. The atom count is 11 for a cystine in mid-chain, but changes to 10 if the cystine joins with another in a disulfide bond. This lead to a much more stable intermolecular interaction.
III) Hydrogen Bonding in water
These hydrogen bonds are at best an interaction, inducing slight positive and negative charges in the Hydrogen and Oxygen/Nitrogen atoms.
The Hydrophilic amino acids have O & N atoms, which form hydrogen bonds with water. These atoms have an uneven distribution of electrons, creating a polar molecule that can interact and form hydrogen bonds with water.
The hydrogen bonds aren't as strong as the covalent bonds in disulfides.
II) Hydrophobic interactions between two leucines
A hydrophobic interaction is formed between two nonpolar molecules.
It describes the preference of nonpolar molecular surfaces to interact with other nonpolar molecular surfaces, thereby displacing water molecules from the interacting surfaces.
A K+ ion and a Cl− ion are directly across from each other on opposite sides of a membrane 7.700 nm thick. What is the electric force on the K+ ion due to the Cl− ion?
Answer:
[tex]-3.896\times 10^{-12} N[/tex] is an electric force on the potassium ion due to the chloride ion.
Explanation:
Charge on potassium ion = [tex]q_1=1.602\times 10^{-19} C[/tex]
Charge on chlorine ion = [tex]=q_2=-1.602\times 10^{-19} C[/tex]
Separation between these two charges = r = [tex]7.700 nm=7.700\times 10^{-9} m[/tex]
[tex]1 nm=10^{-9} m[/tex]
Electric force on the potassium ion due to the chloride ion = F
Coulomb's law is given as ;
[tex]F=K\times \frac{q_1\times q_2}{r^2}[/tex]
[tex]q_1,q_2[/tex] = Charges on both charges
r = distance between the charges
K = Coulomb constant =[tex]9\times 10^{9} N m^2/C^2[/tex]
[tex]F=9\times 10^{9} N m^2/C^2\times \frac{1.602\times 10^{-19} C\times (-1.602\times 10^{-19} C)}{(7.700\times 10^{-9} m)^2}[/tex]
[tex]F=-3.896\times 10^{-12} N[/tex]
(negative sign indicates that attractive force is exerting between two ions)
[tex]-3.896\times 10^{-12} N[/tex] is an electric force on the potassium ion due to the chloride ion.
Final answer:
To find the electric force on a K+ ion due to a Cl- ion, Coulomb's law is used, employing the values for Coulomb's constant, the charge of each ion, and the distance between the ions.
Explanation:
The question refers to calculating the electric force on a K+ ion due to a Cl- ion across a membrane using Coulomb's law. Coulomb's law is defined as F = k * |q1*q2| / r^2, where F is the force between the charges, k is the Coulomb's constant (8.987 x 10^9 N m^2/C^2), q1 and q2 are the magnitudes of the charges (for K+ and Cl-, it's 1.602 x 10^-19 C), and r is the distance between the charges (7.700 nm or 7.700 x 10^-9 m). Plugging these values into Coulomb's law:
F = (8.987 x 10^9) * (1.602 x 10^-19)^2 / (7.700 x 10^-9)^2
After calculating, you will find the electric force exerted on the K+ ion due to the Cl- ion. This principle is crucial in understanding the movement of ions across cell membranes, influencing cell behavior and function.
If 0.500 moles of sulfuric acid and 0.500 moles of Aluminum hydroxide react to make water and aluminum sulfate. a) which reactant is the limiting reactant b) How many moles of Aluminum sulfate is produced?
Answer:
The correct answer to a) is sulfuric acid
b) 1/6 moles of aluminum sulfate is produced
Explanation:
To solve this we need to write out the balaned chemical equation as follows
Al(OH)3(s) + 3 H2SO4(aq) -----> Al2 (SO4)3(aq) + 6 H2O(l)
Here we see that one mole of aluminium hydroxide reacts with three moles of sulphuric acid to form one mole of aluminium sulphate and six ,oles of water
Hence the limiting reactant in this question is the sulphuric acid as 0.5 moles of alumininium hydroxide requires 1.5 moles of sulphuric acid to completely use up the aluminium hydroxide present
Dividing the number of moles of the reactants present by the amount of moles of sulphuric acid required we have
Hence 3 mole of sulphric acid combines with 1 mole of Aluminium hydroxide
0.5 mole of sulfuric acid combines with 0.5÷3 or 1/6 mole of Aluminium hydroxide
Hence the correct answer is sulfuric acid
b) to solve this, since three moles of sulfuric acid produces one mole of aluminium sulfate then 0.5 moles of sulfuric acid produces 0.5/3 or 0.166 mole of aluminum sulfate
hence 1/6 moles of aluminum sulfate is produced